A) \[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\]
B) \[{{A}_{1}}\omega _{1}^{2}={{A}_{2}}\omega _{2}^{2}={{A}_{3}}\omega _{3}^{2}\]
C) \[A_{1}^{2}{{\omega }_{1}}=A_{2}^{2}{{\omega }_{2}}=A_{3}^{2}{{\omega }_{3}}\]
D) \[A_{1}^{2}\sqrt{{{\omega }_{1}}}=A_{2}^{2}\sqrt{{{\omega }_{2}}}=A_{3}^{2}\sqrt{{{\omega }_{3}}}\]
Correct Answer: A
Solution :
\[y=A\sin \omega t\] \[v=\frac{dy}{dt}=A\omega \cos \omega t\] \[\therefore \]Maximum speed \[=A\omega \] Since all the bodies have same maximum speeds, so\[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}\] Hence, the correction option is [a].
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